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Finding potential of a given wavefunction in spherical polar

  1. Nov 9, 2014 #1
    1. The problem statement, all variables and given/known data
    The ground state wavefuntion
    of a system in spherical polar
    coordinates is given by:
    Ψ (r,θ, φ)= (A/r) [exp (-ar) -
    exp (-br)] where a, b, A are
    constants.
    i) Determine A as a function
    of a and b, so as to normalize
    the wavefuntion.
    ii) From Schrödinger equation
    find V (r) in terms of a and b
    iii) From potential behaviour
    find the energy eigenvalue if
    b=6a in the ground state.

    2. Relevant equations
    integral |ψ|^2 dτ=1
    Hψ=Eψ
    H= -hbar/2m grad^2 + V

    3. The attempt at a solution
    I integrated | ψ|^2 r^2 dr
    sinθ dθ dφ =1
    I found A= {1/a-b}[ab* (a
    +b) /2π]^(1/2)
    After putting the wavefuntion
    in time independent
    schrodinger's (Hψ=Eψ)the
    calculation is getting pretty
    elaborate.Cant figure out how
    to find the potential.This is a
    University exam question and
    each of the three question
    carried 2 marks.please help
    me calculate the potential.
     
  2. jcsd
  3. Nov 9, 2014 #2

    Simon Bridge

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    Please show your working when you applied the schrodinger equation.
     
  4. Nov 9, 2014 #3
    let 2m/hbar^2=k
    then -1/k grad^2 ψ +Vψ=Eψ
    or grad^2 ψ =k (V-E)ψ
    given wavefuntion is independent of θ, φ
    so in spherical polar grad^2= (1/r^2)δ/δr (r^2 δ/δr)
    I am getting (r^2 δ/δr)= r [b exp(-br)-a exp (-ar)] - [exp (-ar)-exp (-br)]
    next differenting the above again w.r.t r and multipying with 1/r^2
    grad^2 ψ= ( 1/r) [a^2 exp (-ar)- b^2 exp (-br)]
    'A' depends on a and b which are constants .So I guess A will cancel out.
    So now my TISE looks like
    ( 1/r) [a^2 exp (-ar)- b^2 exp (-br)]=k (V-E) ( 1/r) [exp (-ar)- exp (-br)]
    What should I do after this??
     
    Last edited: Nov 9, 2014
  5. Nov 10, 2014 #4
    Can somebody at all confirm whether it is possible to find the potential?
     
  6. Nov 18, 2014 #5
    utube
     
    Last edited: Nov 18, 2014
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